Difference between revisions of "Compactness"

[ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [ilmath]\implies[/ilmath] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering

Suppose that the space [math](Y,\mathcal{J}_\text{subspace})[/math] is compact and that [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] (where each [math]A_\alpha\in\mathcal{J}[/math] - that is each set is open in [math]X[/math]) is an open covering (which is to say [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath])

Then the collection [math]\{A_\alpha\cap Y|\alpha\in I\}[/math] is a covering of [math]Y[/math] by sets open in [math]Y[/math] (by definition of being a subspace)

By hypothesis [math]Y[/math] is compact, hence a finite sub-collection [math]\{A_{\alpha_i}\cap Y\}^n_{i=1}[/math] covers [math]Y[/math] (as to be compact every open cover must have a finite subcover)

Then [math]\{A_{\alpha_i}\}^n_{i=1}[/math] is a sub-collection of [math]\mathcal{A}[/math] that covers [math]Y[/math].

Proof of details

As The intersection of sets is a subset of each set and [math]\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y[/math] we see

[math]x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y[/math] [math]\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}[/math]

The important part being [math]x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}[/math]

then by the implies and subset relation we have [math]Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}[/math] and conclude [math]Y\subset\cup^n_{i=1}A_{\alpha_i}[/math]

Warning: this next bit looks funny - do not count on!

Lastly, as [math]\mathcal{A}[/math] was a covering [math]\cup_{\alpha\in I}A_\alpha=Y[/math].

It is clear that [math]x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha[/math] so again implies and subset relation we have:

[math]\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y[/math] thus concluding [math]\cup^n_{i=1}A_{\alpha_i}\subset Y[/math]

Combining [math]Y\subset\cup^n_{i=1}A_{\alpha_i}[/math] and [math]\cup^n_{i=1}A_{\alpha_i}\subset Y[/math] we see [math]\cup^n_{i=1}A_{\alpha_i}=Y[/math]

Thus [math]\{A_{\alpha_i}\}^n_{i=1}[/math] is a finite covering of [math]Y[/math] consisting of open sets from [math]X[/math]

End of warning - I've left this here because I must have put it in for a reason!

TODO: What was I hoping to do here?

[ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [math]\impliedby[/math] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering

Suppose that every covering of [math]Y[/math] by sets open in [math]X[/math] contains a finite subcollection covering [math]Y[/math]. We need to show [math]Y[/math] is compact.

Suppose we have a covering, [math]\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}[/math] of [math]Y[/math] by sets open in [math]Y[/math]

For each [math]\alpha[/math] choose an open set [math]A_\alpha[/math] open in [math]X[/math] such that: [math]A'_\alpha=A_\alpha\cap Y[/math]

Then the collection [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] covers [math]Y[/math]

By hypothesis we have a finite sub-collection from [ilmath]\mathcal{A} [/ilmath] of things open in [math]X[/math] that cover [math]Y[/math]

Thus the corresponding finite subcollection of [math]\mathcal{A}'[/math] covers [math]Y[/math]